

A051179


a(n) = 2^(2^n)  1.


29



1, 3, 15, 255, 65535, 4294967295, 18446744073709551615, 340282366920938463463374607431768211455, 115792089237316195423570985008687907853269984665640564039457584007913129639935
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OFFSET

0,2


COMMENTS

In a tree with binary nodes (0, 1 children only), the maximum number of unique child nodes at level n.
Number of binary trees (each vertex has 0, or 1 left, or 1 right, or 2 children) such that all leaves are at level n. Example: a(1) = 3 because we have (i) root with a left child, (ii) root with a right child and (iii) root with two children. a(n) = A000215(n)  2.  Emeric Deutsch, Jan 20 2004
Similarly, this is also the number of full balanced binary trees of height n. (There is an obvious 1to1 correspondence between the two sets of trees.)  David Hobby (hobbyd(AT)newpaltz.edu), May 02 2010
Partial products of A000215.
The first 5 terms n (only) have the property that phi(n)=(n+1)/2, where phi(n) = A000010(n) is Euler's totient function.  Lekraj Beedassy, Feb 12 2007
If A003558(n) is of the form 2^n and A179480(n+1) is even, then (2^(A003558(n)  1) is in the set A051179. Example: A003558(25) = 8 with A179480(25) = 4, even. Then (2^8  1) = 255.  Gary W. Adamson, Aug 20 2012
For any odd positive a(0), the sequence a(n) = a(n1) * (a(n1) + 2) gives a constructive proof that there exist integers with at least n distinct prime factors, e.g. a(n), since omega(a(n)) >= n. As a corollary, this gives a constructive proof of Euclid's theorem stating that there are an infinity of primes.  Daniel Forgues, Mar 07 2017
From Sergey Pavlov, Apr 24 2017: (Start)
I conjecture that, for n > 7, omega(a(n)) > omega(a(n1)) > n.
It seems that the largest prime divisor p(n+1) of a(n+1) is always bigger than the largest prime divisor of a(n): p(n+1) > p(n). For 3 < n < 8, p(n+1) > 100 * p(n).
(End)


REFERENCES

M. Aigner and G. M. Ziegler, Proofs from The Book, SpringerVerlag, Berlin, 1999; see p. 4.


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..13
For rate of growth see A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fib. Quart., 11 (1973), 429437.
Index entries for sequences of form a(n+1)=a(n)^2 + ...


FORMULA

a(n) = A000215(n)  2.
a(n) = (a(n1) + 1)^2  1, a(0) = 1. [ or a(n) = a(n1)(a(n1) + 2) ].
1 = 2/3 + 4/15 + 16/255 + 256/65535 + ... = Sum_{n>=0} A001146(n)/a(n+1) with partial sums: 2/3, 14/15, 254/255, 65534/65535, ...  Gary W. Adamson, Jun 15 2003
a(n) = b(n1) where b(1)=1, b(n) = Product_{k=1..n1} (b(k) + 2).  Benoit Cloitre, Sep 13 2003
A136308(n) = A007088(a(n)).  Jason Kimberley, Dec 19 2012
A000215(n) = a(n+1) / a(n).  Daniel Forgues, Mar 07 2017


EXAMPLE

15 = 3*5; 255 = 3*5*17; 65535 = 3*5*17*257; 4294967295 = 3*5*17*257*65537;
18446744073709551615 = 3*5*17*257*641*65537*6700417;
340282366920938463463374607431768211455 = 3*5*17*257*641*65537*274177*6700417*67280421310721; ...  Daniel Forgues, Mar 07 2017


MAPLE

A051179:=n>2^(2^n)1; seq(A051179(n), n=0..8); # Wesley Ivan Hurt, Feb 08 2014


MATHEMATICA

Table[2^(2^n)1, {n, 0, 9}] (* Vladimir Joseph Stephan Orlovsky, Mar 16 2010 *)


PROG

(PARI) a(n)=if(n<0, 0, 2^2^n1)
(MAGMA) [2^(2^n)1: n in [0..8]]; // Vincenzo Librandi, Jun 20 2011


CROSSREFS

Cf. A001146, A007018.
Cf. A003558, A179480, A000215.
Sequence in context: A247174 A277626 A050474 * A122591 A120607 A013352
Adjacent sequences: A051176 A051177 A051178 * A051180 A051181 A051182


KEYWORD

nonn,easy,nice


AUTHOR

Alan DeKok (aland(AT)ox.org)


STATUS

approved



